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|f                     430                         Moment of Resistance.


1 p                         We may also represent the moment in terms of the limiting

!!                    stress/(sometimes/c, and sometimes/t).    Then:

if Jj                                       Bending moment = moment of resistance

j|                                                        Bm=/Z     ........................ (4)


I/                    and Z is known as the modulus of section,*


||                           Let y = distance of furthest fibre from axis:

I                     By (i)   /=           by (3) Bm =       and by (4)    Bm=/Z

Z-        .-.   Z = *...... (5)

p                   P                                y

and    Bm=/-                               .

_____I                              (Seej). 1079.)

The value of Z can now be found; thus,


% f

r                       i                                 rr,      r "   '"        ^          f b frfi

rectangular Sections       /Z =/------r -   =/-

and for circular sections    /Z = / -r- -

--------------------------------      J     64         2,

Graphic   Solution   for   Moment   of   Resistance.

Taking, first, a rectangular section (Fig. 384), draw the neutral
axis A B. Then c D will be the line of limiting (or greatest) stress,
and the value of any horizontal fibre E F to resist stress will be
found by projecting to c D and joining c D to N, thus obtaining
the intercept G M. Every fibre being thus treated, the sum of
the -virtual stress areas will be the areas c{# N and H j N, which
each make one force of the couple when multiplied by the
limiting stress f. K and L are the centres of gravity of the areas.

Moment of resistance (generally) = one force x total arm
Moment of rectangular section     = f\area C D N [ x arm K L

Unsymmetrical sections are treated at Figs. 389, 390 by this
method, which can be applied to any section. (SeeApp.IL>p. 847.)

* Z = virtual area x arm.    (See Figs. 384, 385.)